DP Maths AA SL - Statistics and Probability - Probability and Probability Distributions

DP Maths AA SL - Statistics and Probability - Probability and Probability Distributions - Full notes

RevisionPrep

DP Maths AA SL | Probability and Probability Distributions

Probability and Probability Distributions

This topic contains 3 notes:

4.5 Probability
4.7 Discrete Random Variables
4.9 Normal Distribution

DP MATHEMATICS AA · TOPIC 4: STATISTICS AND PROBABILITY

4.5

Probability

Full notes for SL · Syllabus SL 4.5, 4.6, 4.11

Contents

1	Experiments, outcomes and events
2	Theoretical and experimental probability
3	Sample space diagrams and expected numbers
4	Venn diagrams and the addition rule
5	Mutually exclusive events
6	Conditional probability
7	Independent events
8	Tree diagrams and sampling without replacement
9	Sample spaces: listing systematically
10	Venn diagrams with three sets
11	Conditional probability from tables
12	Independence: testing and using
13	Tree diagrams with three stages
14	Without replacement and dependent events
15	Conditional probability from trees
16	Probability and counting
17	Expected numbers and simulation
18	Mixed review: worked problems
19	Additional exam-style practice
20	Common pitfalls in probability
21	Probability distributions from experiments
22	Exam-style questions with markscheme
23	Common misconceptions and links
24	Check your understanding
25	Summary and topic map
26	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
use the language of trials, outcomes, sample spaces and events
find theoretical and experimental probabilities, and expected numbers of occurrences
use sample space diagrams, tables and Venn diagrams
use the addition rule P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
identify mutually exclusive events
calculate and interpret conditional probability P(A \| B)
test for independence using P(A ∩ B) = P(A)P(B)
use tree diagrams, including sampling without replacement

This booklet covers SL 4.5, 4.6, 4.11, which is common to SL and HL.

Guiding questions

How can uncertainty be measured and combined? How does knowing that one event has happened change the probability of another?

How this topic is examined

Probability is tested on both papers: Venn diagrams and tables on Paper 1, conditional probability and independence 'show that' questions, and tree diagrams for multi-stage experiments. Conditional probability from a tree is a very common exam task.

Define events clearly and write probability statements in notation, for example P(A | B), before calculating.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Experiments, outcomes and events

The vocabulary of chance.

Think about it

A weather forecast says there is a 70% chance of rain tomorrow. What does that number actually mean?

A trial is one performance of an experiment (rolling a die). An outcome is a possible result. The sample space U is the set of all possible outcomes. An event A is a set of outcomes. The complement A′ is the event 'A does not happen'.

2 Theoretical and experimental probability

Equally likely outcomes and relative frequency.

If all outcomes are equally likely, P(A) = n(A)/n(U). If not, probability is estimated by relative frequency: P(A) ≈ (number of times A occurred)/(number of trials); this estimate improves as the number of trials increases.

The expected number of occurrences of A in n trials is np, where p = P(A).

WORKED EXAMPLE 1 Expected number (basic)

A biased coin landed heads 58 times in 200 tosses. Estimate P(heads) and the expected number of heads in 1000 tosses.

Solution

P(H) ≈ 58/200 = 0.29A1

Expected number = 1000 × 0.29 = 290A1

Practice exercise 1

1. A card is drawn from a standard pack of 52. Find P(a king or a heart).

2. The probability that a seed germinates is 0.85. How many of 400 seeds are expected to germinate?

3 Sample space diagrams and expected numbers

List or tabulate all outcomes.

For two-stage experiments with equally likely outcomes, a sample space diagram (a grid) lists all combinations. Two fair dice give 36 equally likely outcomes.

+	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

WORKED EXAMPLE 2 Two dice (standard)

Two fair dice are rolled. Find P(sum = 8) and P(sum > 9).

Solution

Sum 8: (2,6), (3,5), (4,4), (5,3), (6,2): 5/36A1

Sum > 9: sums 10, 11, 12: 3 + 2 + 1 = 6 outcomes: 1/6A1

Practice exercise 2

1. Two dice are rolled. Find P(at least one six).

2. Two dice are rolled. Find P(the product is even).

4 Venn diagrams and the addition rule

Count each region once.

A Venn diagram shows events as overlapping circles inside the rectangle U. A ∩ B (A and B) is the overlap; A ∪ B (A or B, or both) is everything in either circle. Adding P(A) and P(B) counts the overlap twice, so:

Figure 1. 40 students: M = studies Mathematics HL, P = studies Physics.

WORKED EXAMPLE 3 Reading a Venn diagram (standard)

Use the Venn diagram. Find (a) P(M); (b) P(M ∪ P); (c) P(M′ ∩ P).

Solution

(a) (12 + 8)/40 = 1/2A1

(b) 35/40 = 7/8A1

Practice exercise 3

1. P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.7. Find P(A ∩ B).

2. In a class of 30, 18 play football, 12 play tennis and 5 play neither. Find P(plays both).

5 Mutually exclusive events

Cannot happen together.

Events A and B are mutually exclusive if they cannot both occur: P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B). On a Venn diagram the circles do not overlap. 'Rolling a 2' and 'rolling an odd number' are mutually exclusive.

6 Conditional probability

Restrict the sample space.

The conditional probability of A given B, P(A | B), is the probability of A when we know B has occurred. Knowing B restricts the sample space to B:

WORKED EXAMPLE 4 From a Venn diagram (standard)

Using the Venn diagram, find P(M | P).

Solution

Restrict to P: 8 + 15 = 23 studentsM1

P(M | P) = 8/23A1

WORKED EXAMPLE 5 From a two-way table (standard)

In a survey of 200 people, 80 are under 30; of these, 60 use social media daily. Of those 30 or over, 50 use it daily. Find P(under 30 | uses daily).

Solution

Daily users: 60 + 50 = 110M1

P = 60/110 = 6/11A1

Common mistake

Confusing P(A | B) with P(B | A). The event after the bar is the one known to have occurred.

Practice exercise 4

1. P(A ∩ B) = 0.12 and P(B) = 0.3. Find P(A | B).

2. Two dice are rolled. Given the sum is 8, find P(at least one 6).

3. P(A) = 0.6, P(B | A) = 0.5, P(B | A′) = 0.2. Find P(B).

7 Independent events

One does not affect the other.

Events A and B are independent if the occurrence of one does not change the probability of the other: P(A | B) = P(A). Equivalently:

To test for independence, calculate P(A)P(B) and compare it with P(A ∩ B). Do not confuse independent with mutually exclusive: two events with non-zero probabilities that are mutually exclusive cannot be independent, since P(A ∩ B) = 0 ≠ P(A)P(B).

WORKED EXAMPLE 6 Testing independence (exam level)

Use the Venn diagram to determine whether M and P are independent.

Solution

P(M) = 20/40 = 0.5, P(P) = 23/40 = 0.575; product 0.2875M1

P(M ∩ P) = 8/40 = 0.2A1

0.2 ≠ 0.2875, so not independentR1

Practice exercise 5

1. P(A) = 0.4, P(B) = 0.5, P(A ∪ B) = 0.7. Are A and B independent?

2. A and B are independent with P(A) = 0.3 and P(B) = 0.6. Find P(A ∪ B).

8 Tree diagrams and sampling without replacement

Multiply along branches, add between branches.

A tree diagram shows multi-stage experiments. Probabilities on each set of branches from a point add to 1. Multiply along a path to find the probability of that combined outcome; add the probabilities of all paths that satisfy the event.

When sampling without replacement, the second-stage probabilities change because the population has changed.

Figure 2. Two counters drawn without replacement from a bag of 4 red and 6 blue.

WORKED EXAMPLE 7 Without replacement (standard)

Two counters are drawn without replacement from a bag of 4 red and 6 blue. Find P(one of each colour).

Solution

P(RB) + P(BR) = (4/10)(6/9) + (6/10)(4/9)M1

= 24/90 + 24/90 = 8/15A1

WORKED EXAMPLE 8 Conditional from a tree (exam level)

Using the same experiment, find P(first is red | second is red).

Solution

P(second R) = (4/10)(3/9) + (6/10)(4/9) = 12/90 + 24/90 = 36/90M1 A1

P(first R and second R) = 12/90A1

P = (12/90)/(36/90) = 1/3A1

Practice exercise 6

1. A test for a condition is positive for 95% of people with it and 8% of people without it. 2% of people have the condition. Find P(positive).

2. Using the previous question, find P(has condition | positive).

3. Three cards are drawn without replacement from 5 red and 3 black. Find P(all red).

9 Sample spaces: listing systematically

Do not miss outcomes.

List outcomes systematically (in order, or in a table or tree) so that none is missed or counted twice. For two independent stages, the sample space has (outcomes of stage 1) × (outcomes of stage 2) equally likely outcomes when each stage is fair.

WORKED EXAMPLE 9 Coin and die (basic)

A coin is tossed and a die rolled. Find P(head and an even number).

Solution

12 equally likely outcomes; favourable 3M1

1/4A1

WORKED EXAMPLE 10 Spinners (standard)

Two spinners show 1–4 and 1–3. Find P(the product is even).

Solution

12 outcomes; odd products need both odd: 2 × 2 = 4M1

P(even) = 1 − 4/12 = 2/3A1

Practice exercise 7

1. Three coins are tossed. Find P(exactly two heads).

2. Two dice are rolled. Find P(the difference is 2).

10 Venn diagrams with three sets

Fill from the centre outwards.

With three events, fill the triple overlap first, then the pairwise overlaps, then the single regions, and finally the outside. Check that all regions sum to the total.

WORKED EXAMPLE 11 Three sets (exam level)

In a group of 60: 30 study French, 25 Spanish, 20 German; 10 study French and Spanish, 8 French and German, 6 Spanish and German, 3 all three. Find the number studying none.

Solution

Union = 30 + 25 + 20 − 10 − 8 − 6 + 3 = 54M1 A1

None = 6A1

Practice exercise 8

1. Find P(exactly one language) for the group above.

11 Conditional probability from tables

Restrict to the given row or column.

In a two-way table, P(A | B) is found by restricting to the row or column for B and dividing the count for A within it by the total of B.

	Smoker	Non-smoker	Total
Male	35	115	150
Female	25	125	150
Total	60	240	300

WORKED EXAMPLE 12 From a table (standard)

Find P(smoker | male) and P(male | smoker).

Solution

35/150 = 0.233A1

35/60 = 0.583A1

Practice exercise 9

1. Are 'male' and 'smoker' independent? Justify.

12 Independence: testing and using

P(A ∩ B) = P(A)P(B).

To test independence, compare P(A ∩ B) with P(A)P(B), or P(A | B) with P(A). To use independence, multiply probabilities of the separate events. If A and B are independent, so are A and B′, A′ and B, and A′ and B′.

WORKED EXAMPLE 13 Using independence (exam level)

P(A) = 0.6, P(B) = 0.3, A and B independent. Find P(A ∪ B) and P(exactly one occurs).

Solution

P(A ∩ B) = 0.18; P(A ∪ B) = 0.72M1 A1

Exactly one = 0.72 − 0.18 = 0.54A1

Practice exercise 10

1. A and B are independent with P(A ∪ B) = 0.8 and P(A) = 0.5. Find P(B).

13 Tree diagrams with three stages

Multiply along paths.

For three stages, each path has three branches. Multiply along paths and add the paths that satisfy the event. Using complements ('at least one') often saves work.

WORKED EXAMPLE 14 Three shots (exam level)

A player scores with probability 0.7 each shot, independently. Find P(exactly two goals in three shots).

Solution

3 paths, each 0.72(0.3)M1

= 0.441A1

Practice exercise 11

1. Find P(at least one goal in three shots).

2. Find P(the first goal comes on the third shot).

14 Without replacement and dependent events

Probabilities change.

When items are not replaced, later probabilities depend on earlier outcomes. Adjust the numbers at each stage of the tree.

WORKED EXAMPLE 15 Three draws (exam level)

A bag has 5 red and 3 blue balls. Three are drawn without replacement. Find P(exactly two red).

Solution

RRB, RBR, BRR each (5 × 4 × 3)/(8 × 7 × 6)M1

3 × 60/336 = 15/28A1

Practice exercise 12

1. Find P(all three red).

2. Find P(at least one blue).

15 Conditional probability from trees

Reverse the condition.

P(first event | final outcome) = P(path)/P(final outcome). Find the denominator by adding all paths that lead to the final outcome.

WORKED EXAMPLE 16 Reversing (exam level)

In the three-draw example, given that exactly two red balls were drawn, find P(the first ball was blue).

Solution

P(BRR) = 60/336A1

(60/336)/(180/336) = 1/3M1 A1

Practice exercise 13

1. A test is 95% accurate for diseased people and 90% for healthy people; 4% have the disease. Find P(disease | positive).

16 Probability and counting

Combinations for equally likely selections.

When selections are equally likely, P = (favourable selections)/(total selections), using nCr for both. This is often quicker than a tree for several draws.

WORKED EXAMPLE 17 Committee (standard)

A committee of 4 is chosen from 6 men and 5 women. Find P(exactly 2 women).

Solution

6C2 × 5C2/11C4M1

= 150/330 = 5/11A1

Practice exercise 14

1. Five cards are drawn from a pack. Find P(all hearts).

17 Expected numbers and simulation

Long-run frequencies.

The expected number of occurrences of an event in n trials is np. A simulation with random numbers estimates probabilities that are hard to calculate; more trials give better estimates.

WORKED EXAMPLE 18 Expected number (standard)

P(a component fails) = 0.03. In a batch of 400, how many failures are expected?

Solution

400 × 0.03M1

= 12A1

Practice exercise 15

1. A spinner lands on red with probability 0.35. How many reds are expected in 60 spins?

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 19 Mixed problem (exam level)

P(A) = 0.5, P(B) = 0.4, P(A | B) = 0.3. Find P(A ∩ B), P(A ∪ B) and P(B | A).

Solution

P(A ∩ B) = 0.12A1

P(A ∪ B) = 0.78A1

P(B | A) = 0.12/0.5 = 0.24A1

Practice exercise 16

1. Events A and B are mutually exclusive with P(A) = 0.3 and P(B) = 0.45. Find P(A′ ∩ B′).

19 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 20 Paper 2 question (exam level)

Bag A has 3 red and 2 green balls; bag B has 1 red and 4 green. A die is rolled: a 6 means bag A is chosen, otherwise bag B. One ball is drawn. (a) Find P(red). (b) Given that the ball is red, find P(bag A).

Solution

(a) (1/6)(3/5) + (5/6)(1/5) = 3/30 + 5/30 = 4/15M1 A1

(b) (3/30)/(8/30) = 3/8M1 A1

Practice exercise 17

1. Find P(green) in the example.

20 Common pitfalls in probability

Check your reasoning.

Confusing 'and' with 'or'; adding probabilities of events that are not mutually exclusive; multiplying probabilities of events that are not independent; forgetting that order matters on trees; mixing up P(A | B) and P(B | A). A quick check: all probabilities must lie between 0 and 1, and the probabilities of all outcomes must sum to 1.

WORKED EXAMPLE 21 Spot the error (standard)

A student says P(at least one six in two rolls) = 1/6 + 1/6 = 1/3. Correct this.

Solution

The events overlap (both sixes)R1

1 − (5/6)2 = 11/36A1

21 Probability distributions from experiments

Linking to random variables.

Listing the outcomes of an experiment with their probabilities produces a probability distribution of a random variable (4.7). For example, the number of heads in three coin tosses takes values 0, 1, 2, 3 with probabilities 1/8, 3/8, 3/8, 1/8.

WORKED EXAMPLE 22 Distribution (standard)

Find the distribution of the larger score when two dice are rolled.

Solution

P(max = k) = (2k − 1)/36 for k = 1, ..., 6M1 A1

Practice exercise 18

1. Find the distribution of the number of sixes in two rolls of a die.

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 15 minutes)

Events A and B have P(A) = 0.4, P(B) = 0.5 and P(A ∪ B) = 0.7.

(a) Find P(A ∩ B). [2]

(b) Show that A and B are independent. [2]

(d) Draw a Venn diagram showing the probability in each region. [2]

Markscheme

Part	Answer	Marks
(a)	0.4 + 0.5 − 0.7	M1
	0.2	A1
(b)	P(A)P(B) = 0.2	A1
	= P(A ∩ B), so independent	R1 AG
(c)	P(A ∩ B′) = 0.2, P(B′) = 0.5	M1
	0.4	A1
(d)	regions 0.2, 0.2, 0.3	A1
	outside 0.3	A1

Examiner's comments. Part (c) equals P(A), as expected for independent events.

Question 2 (Paper 2 style) (about 12 minutes)

Factory machines X, Y and Z produce 50%, 30% and 20% of all components. The defect rates are 2%, 3% and 5% respectively.

(a) Find the probability that a randomly chosen component is defective. [3]

(b) Given that a component is defective, find the probability that it was made by machine Z. [3]

Markscheme

Part	Answer	Marks
(a)	0.5(0.02) + 0.3(0.03) + 0.2(0.05)	M1 A1
	0.029	A1
(b)	P(Z ∩ D) = 0.01	A1
	0.01/0.029	M1
	0.345	A1

Examiner's comments. A tree diagram makes both parts straightforward.

23 Common misconceptions and links

Students often think…	Actually…
mutually exclusive means independent	they are different ideas
P(A \| B) = P(B \| A)	not in general
add along branches	multiply along, add between
probabilities can exceed 1	0 ≤ P ≤ 1

How 4.5 connects to the rest of the course

Topic	Connection
4.7 Discrete random variables	probability distributions
4.13 Bayes' theorem (HL)	reversing conditional probability
1.10 Counting (HL)	counting outcomes
4.1 Statistics	relative frequency

24 Check your understanding

Section A: multiple choice

1. P(A′) =

A. 1 − P(A) B. P(A) C. 0 D. 1 + P(A)

2. For independent events, P(A ∩ B) =

A. P(A) + P(B) B. 0 C. P(A)P(B) D. P(A | B)

3. Mutually exclusive means

A. P(A ∩ B) = P(A)P(B) B. P(A ∩ B) = 0 C. P(A) = P(B) D. P(A ∪ B) = 1

4. P(A | B) =

A. P(A ∩ B)/P(A) B. P(A)P(B) C. P(B)/P(A) D. P(A ∩ B)/P(B)

5. Two dice: P(double) =

A. 1/36 B. 1/6 C. 1/3 D. 1/12

Section B: structured

6. Two cards are drawn without replacement from a standard pack. Find P(both aces). [2]

7. Show that if A and B are independent, then A and B′ are independent. [3]

Answers

1. A 2. C 3. B 4. D 5. B

6. (4/52)(3/51) = 1/221.

7. P(A ∩ B′) = P(A) − P(A)P(B) = P(A)(1 − P(B)) = P(A)P(B′).

25 Summary and topic map

Idea	Remember
Basics	0 ≤ P(A) ≤ 1; P(A′) = 1 − P(A)
Expected number	np
Addition rule	P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Mutually exclusive	P(A ∩ B) = 0
Conditional	P(A \| B) = P(A ∩ B)/P(B)
Independent	P(A ∩ B) = P(A)P(B)
Trees	multiply along, add between

Before the exam, make sure you…

•define events
•use the correct conditional
•test independence numerically
•adjust probabilities without replacement

Probability on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 3. Topic map for Probability.

26 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. 16/52 = 4/13

2. 340

Practice exercise 2

1. 11/36

2. 27/36 = 3/4

Practice exercise 3

1. 0.2

2. 18 + 12 − 25 = 5: 1/6

Practice exercise 4

1. 0.4

2. 2/5

3. 0.3 + 0.08 = 0.38

Practice exercise 5

1. P(A ∩ B) = 0.2 = 0.4 × 0.5: yes

2. 0.3 + 0.6 − 0.18 = 0.72

Practice exercise 6

1. 0.019 + 0.0784 = 0.0974

2. 0.019/0.0974 = 0.195

3. (5/8)(4/7)(3/6) = 5/28

Practice exercise 7

1. 3/8

2. 8/36 = 2/9

Practice exercise 8

1. only F 15, only S 12, only G 9: 36/60 = 0.6

Practice exercise 9

1. P(smoker) = 0.2 ≠ 0.233: not independent

Practice exercise 10

1. 0.5 + p − 0.5p = 0.8: p = 0.6

Practice exercise 11

1. 1 − 0.027 = 0.973

2. 0.3 × 0.3 × 0.7 = 0.063

Practice exercise 12

1. 60/336 = 5/28

2. 23/28

Practice exercise 13

1. 0.038/(0.038 + 0.096) = 0.284

Practice exercise 14

1. 13C5/52C5 = 0.000495

Practice exercise 15

1. 21

Practice exercise 16

1. 1 − 0.75 = 0.25

Practice exercise 17

1. 11/15

Practice exercise 18

1. 0: 25/36; 1: 10/36; 2: 1/36

DP MATHEMATICS AA · TOPIC 4: STATISTICS AND PROBABILITY

4.7

Discrete Random Variables and the Binomial Distribution

Full notes for SL · Syllabus SL 4.7, 4.8

Contents

1	Discrete random variables
2	Probability distributions
3	Expected value
4	Fair games and applications of expectation
5	The binomial distribution
6	Calculating binomial probabilities
7	Mean and variance of the binomial distribution
8	Problems with unknown n or p
9	Building a probability distribution
10	Expected value calculations
11	Variance of a discrete random variable
12	Unknown probabilities from conditions
13	Games and fairness
14	The binomial distribution: recognizing it
15	Binomial calculations: exact and with technology
16	Mean and variance of the binomial
17	Binomial in context: decisions
18	Mixed review: worked problems
19	Additional exam-style practice
20	Linear transformations of random variables (extension)
21	Exam-style questions with markscheme
22	Common misconceptions and links
23	Check your understanding
24	Summary and topic map
25	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
define a discrete random variable and its probability distribution
use the condition ΣP(X = x) = 1 to find unknowns
calculate expected value E(X) and interpret it, including fair games
state the conditions for a binomial distribution
calculate binomial probabilities with the formula and with technology
use E(X) = np and Var(X) = np(1 − p)
solve problems with unknown n or p

This booklet covers SL 4.7, 4.8, which is common to SL and HL.

Guiding questions

How can we describe the long-run behaviour of a random process? When is the binomial distribution an appropriate model?

How this topic is examined

Discrete distributions and the binomial appear on both papers: completing a probability table, finding E(X), deciding whether a game is fair, and binomial probabilities (including 'at least' and 'at most'). Paper 2 uses the GDC binomial pdf and cdf.

Define the random variable in words and state its distribution, for example X ~ B(10, 0.3), before calculating.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Discrete random variables

Numbers from random outcomes.

Think about it

You play a game many times. How can you predict your average winnings per game without playing at all?

A random variable X assigns a number to each outcome of an experiment. It is discrete if it takes a countable set of values, such as the number of heads in four tosses (0, 1, 2, 3, 4). Capital X denotes the variable and small x a particular value; P(X = x) is the probability that X takes the value x.

2 Probability distributions

All values and their probabilities.

A probability distribution lists every possible value of X with its probability, in a table, a formula or a graph. The probabilities must satisfy:

WORKED EXAMPLE 1 Finding an unknown (standard)

P(X = x) = k(x + 1) for x = 0, 1, 2, 3. Find k and P(X ≥ 2).

Solution

k(1 + 2 + 3 + 4) = 1, so k = 1/10M1 A1

P(X ≥ 2) = 3/10 + 4/10 = 7/10A1

x	0	1	2	3
P(X = x)	0.1	0.2	0.3	0.4

Practice exercise 1

1. The table gives P(X = x) = 0.2, 0.35, k, 0.15 for x = 1, 2, 3, 4. Find k and P(X < 3).

2. A die is biased so that P(6) = 0.3 and the other faces are equally likely. Find P(1).

3 Expected value

The long-run average.

The expected value (mean) E(X) is the average value of X over many repetitions. It need not be a possible value of X: the expected value of a fair die is 3.5.

WORKED EXAMPLE 2 Expected value (basic)

Find E(X) for the distribution in the table above.

Solution

0(0.1) + 1(0.2) + 2(0.3) + 3(0.4)M1

= 2A1

WORKED EXAMPLE 3 Unknowns from mean (exam level)

X takes values 1, 2, 3 with probabilities a, 0.5 and b. Given E(X) = 2.2, find a and b.

Solution

a + b = 0.5 and a + 1 + 3b = 2.2M1

a + 3b = 1.2; subtract: 2b = 0.7A1

b = 0.35, a = 0.15A1

Practice exercise 2

1. Find E(X) for a fair die.

2. X has values −2, 0, 5 with probabilities 0.5, 0.3, 0.2. Find E(X).

4 Fair games and applications of expectation

Expected gain.

In a game, let X be the player's gain (winnings minus cost). The game is fair if E(X) = 0. If E(X) < 0, the player loses on average and the organizer profits. Expected values also give expected costs, profits or numbers in business and insurance contexts.

WORKED EXAMPLE 4 A dice game (exam level)

A player pays $3 to roll two dice. They win $10 for a double and $2 if the sum is 7, otherwise nothing. Find the expected gain and decide whether the game is fair.

Solution

P(double) = 6/36, P(sum 7) = 6/36A1

Expected winnings = 10(1/6) + 2(1/6) = $2M1 A1

Expected gain = 2 − 3 = −$1: not fair, favours the organizerR1

Practice exercise 3

1. A raffle sells 500 tickets at $2, with one $300 prize and five $50 prizes. Find the expected gain per ticket.

2. Find the prize for a double in the dice game above that makes the game fair.

5 The binomial distribution

Counting successes in fixed trials.

A random variable X follows a binomial distribution, X ~ B(n, p), when: there is a fixed number n of trials; each trial has two outcomes (success or failure); the probability of success p is constant; and the trials are independent. X counts the number of successes.

Examples: the number of heads in 10 tosses; the number of defective items in a random sample (from a large batch); the number of correct guesses on a multiple-choice test.

Why it works

Any particular sequence with r successes and n − r failures has probability pr(1 − p)n−r, by independence. The number of such sequences is the number of ways of choosing which r of the n trials are successes, nCr. Adding the probabilities of all these sequences gives the formula.

Figure 1. The distribution of X ~ B(10, 0.3).

6 Calculating binomial probabilities

Formula on Paper 1, GDC on Paper 2.

Use binomial pdf for P(X = r) and binomial cdf for P(X ≤ r). Convert other inequalities: P(X < r) = P(X ≤ r − 1); P(X ≥ r) = 1 − P(X ≤ r − 1); P(X > r) = 1 − P(X ≤ r).

WORKED EXAMPLE 5 Exact calculation (standard)

X ~ B(5, 0.4). Find P(X = 2) exactly.

Solution

5C2 (0.4)2(0.6)3 = 10 × 0.16 × 0.216M1

= 0.3456A1

WORKED EXAMPLE 6 Using the GDC (standard)

X ~ B(20, 0.15). Find (a) P(X ≤ 3); (b) P(X ≥ 5); (c) P(2 < X ≤ 6).

Solution

(a) cdf: 0.648A1

(b) 1 − P(X ≤ 4) = 0.170A1

Common mistake

Writing P(X ≥ 5) = 1 − P(X ≤ 5). The correct complement is 1 − P(X ≤ 4).

Practice exercise 4

1. X ~ B(8, 0.5). Find P(X = 4) exactly.

2. X ~ B(12, 0.25). Find P(X < 3) and P(X > 4).

3. Twenty seeds each germinate with probability 0.8. Find the probability that at least 18 germinate.

7 Mean and variance of the binomial distribution

np and np(1 − p).

WORKED EXAMPLE 7 Mean and variance (standard)

A multiple-choice test has 30 questions with 4 options each. A student guesses every answer. Find the mean and standard deviation of the number correct.

Solution

X ~ B(30, 0.25)A1

E(X) = 7.5; Var(X) = 30(0.25)(0.75) = 5.625A1

SD = 2.37A1

Practice exercise 5

1. X ~ B(n, p) with mean 6 and variance 4.2. Find n and p.

2. Find the mode of X ~ B(10, 0.3) from the graph above.

8 Problems with unknown n or p

Use inequalities or equations.

WORKED EXAMPLE 8 Smallest n (exam level)

The probability that a student is left-handed is 0.12. Find the smallest group size n for which the probability of at least one left-hander exceeds 0.9.

Solution

1 − 0.88n > 0.9, so 0.88n < 0.1M1

n > ln 0.1/ln 0.88 = 18.01A1

n = 19A1

WORKED EXAMPLE 9 Finding p (exam level)

X ~ B(6, p) and P(X = 0) = 0.000729. Find p.

Solution

(1 − p)6 = 0.000729M1

1 − p = 0.3A1

p = 0.7A1

Practice exercise 6

1. How many times must a fair die be rolled so that P(at least one six) > 0.99?

2. X ~ B(n, 0.5) and P(X = 0) = 1/64. Find n.

9 Building a probability distribution

From an experiment to a table.

Identify all possible values of X, find the probability of each (using sample spaces, trees or counting), and check that the probabilities sum to 1.

WORKED EXAMPLE 10 Two dice (standard)

X is the absolute difference of two dice scores. Find the distribution of X.

Solution

Count outcomes: 0: 6, 1: 10, 2: 8, 3: 6, 4: 4, 5: 2M1 A1

P(X = x) = 6/36, 10/36, 8/36, 6/36, 4/36, 2/36A1

Practice exercise 7

1. X is the number of heads when 3 coins are tossed. Find its distribution.

2. Find E(X) for the dice difference.

10 Expected value calculations

ΣxP(X = x).

E(X) is a weighted average of the values of X. It may not be a possible value of X. For functions of X, E(g(X)) = Σg(x)P(X = x).

WORKED EXAMPLE 11 Expected value of a function of X (standard)

For P(X = 0, 1, 2, 3) = 0.2, 0.3, 0.4, 0.1, find E(X) and E(X2).

Solution

E(X) = 0.3 + 0.8 + 0.3 = 1.4A1

E(X2) = 0.3 + 1.6 + 0.9 = 2.8A1

Practice exercise 8

1. Find E(2X + 1) for the distribution above.

2. Find E(1/(X + 1)).

11 Variance of a discrete random variable

Spread about the mean.

Var(X) = E(X2) − (E(X))2; the standard deviation is its square root. (Variance is formally in the HL syllabus; SL students use technology for the standard deviation.)

WORKED EXAMPLE 12 Variance (standard)

Find Var(X) for the distribution above.

Solution

2.8 − 1.96M1

= 0.84A1

Practice exercise 9

1. Find the variance of the score on a fair die.

12 Unknown probabilities from conditions

Two equations, two unknowns.

Typical conditions: probabilities sum to 1, and a given mean. Each gives a linear equation in the unknowns.

WORKED EXAMPLE 13 Two unknowns (exam level)

X takes values 1, 2, 3, 4 with probabilities 0.1, a, b, 0.2. E(X) = 2.6. Find a and b.

Solution

a + b = 0.7; 0.1 + 2a + 3b + 0.8 = 2.6M1

2a + 3b = 1.7: b = 0.3, a = 0.4A1 A1

Practice exercise 10

1. P(X = x) = k(x + 2) for x = 1, 2, 3. Find k and E(X).

13 Games and fairness

Expected gain.

Define X as the gain (prize minus cost). The game is fair if E(X) = 0. Find the cost (or a prize) that makes a game fair by solving E(X) = 0.

WORKED EXAMPLE 14 Fair cost (exam level)

A player rolls two dice. They win $12 for a double six, $3 for any other double, nothing otherwise. Find the fair entry fee.

Solution

E(winnings) = 12(1/36) + 3(5/36) = 27/36M1 A1

Fair fee $0.75A1

Practice exercise 11

1. A card is drawn from a pack. An ace wins $10, a picture card $5, otherwise nothing. Find the fair price.

14 The binomial distribution: recognizing it

Four conditions.

X ~ B(n, p) when there are n independent trials, each with the same probability p of success, and X counts the successes. Check every condition in context.

WORKED EXAMPLE 15 Recognizing (standard)

Is the number of red cards in 5 draws with replacement from a pack binomial?

Solution

Yes: n = 5, p = 1/2, independent with replacement: B(5, 0.5)R1 A1

Practice exercise 12

1. Is the number of girls in a randomly chosen family of 4 children binomial? State any assumption.

15 Binomial calculations: exact and with technology

pdf and cdf.

P(X = r) = nCr pr(1 − p)n−r. On Paper 2, use binompdf and binomcdf, converting inequalities: P(X ≥ r) = 1 − P(X ≤ r − 1).

WORKED EXAMPLE 16 Exact (standard)

X ~ B(6, 0.3). Find P(X = 2) exactly.

Solution

15(0.09)(0.7)4M1

= 0.324A1

WORKED EXAMPLE 17 Technology (standard)

X ~ B(25, 0.15). Find P(X ≥ 5) and P(2 ≤ X ≤ 6).

Solution

1 − cdf(4) = 0.318A1

cdf(6) − cdf(1) = 0.837M1 A1

Practice exercise 13

1. X ~ B(10, 0.4). Find P(X < 3).

2. X ~ B(20, 0.9). Find P(X = 20) and P(X ≤ 17).

16 Mean and variance of the binomial

np and np(1 − p).

E(X) = np and Var(X) = np(1 − p). Given the mean and variance, solve for n and p: 1 − p = Var/mean.

WORKED EXAMPLE 18 Finding n and p (exam level)

X ~ B(n, p) with mean 4.8 and variance 2.88. Find n and p.

Solution

1 − p = 2.88/4.8 = 0.6M1

p = 0.4, n = 12A1 A1

Practice exercise 14

1. Find the mean and SD of B(50, 0.08).

17 Binomial in context: decisions

Using probabilities to decide.

Probabilities from a binomial model can support decisions: whether a result is unusual, whether a supplier's claim is believable, or how many items to test.

WORKED EXAMPLE 19 Unusual result (exam level)

A supplier claims 5% of items are faulty. In a sample of 40, 6 are faulty. Is this unusual if the claim is true?

Solution

P(X ≥ 6), X ~ B(40, 0.05) = 0.0139M1 A1

Very unlikely (under 2%): evidence against the claimR1

Practice exercise 15

1. How many items must be tested for P(at least one faulty) > 0.9 if p = 0.05?

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 20 Mixed problem (exam level)

A game: roll a die until a six appears, at most 3 rolls. X is the number of rolls. (a) Find the distribution of X. (b) Find E(X).

Solution

(a) P(1) = 1/6; P(2) = 5/36; P(3) = 25/36A1 A1

(b) 1/6 + 10/36 + 75/36 = 91/36 ≈ 2.53M1 A1

Practice exercise 16

1. X ~ B(8, 0.25). Find the most likely value of X.

19 Additional exam-style practice

Paper 2.

WORKED EXAMPLE 21 Paper 2 question (exam level)

70% of customers buy a drink. On a day with 30 customers: (a) find the expected number buying a drink; (b) find P(more than 25 buy); (c) find P(at most 18 buy).

Solution

(a) 21A1

(b) 1 − cdf(25) = 0.0302M1 A1

Practice exercise 17

1. Over 5 days, find P(more than 25 buy on at least one day).

20 Linear transformations of random variables (extension)

E(aX + b) and Var(aX + b).

E(aX + b) = aE(X) + b and Var(aX + b) = a2Var(X). Profits, costs and unit conversions are linear transformations of random variables.

WORKED EXAMPLE 22 Profit (standard)

Daily sales X have E(X) = 40 and Var(X) = 25. Profit P = 3X − 50. Find E(P) and the SD of P.

Solution

E(P) = 70A1

Var(P) = 9 × 25 = 225: SD 15A1

Practice exercise 18

1. E(X) = 5, Var(X) = 2. Find E(10 − 2X) and Var(10 − 2X).

21 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 12 minutes)

A random variable X has the distribution P(X = x) = (x + k)/20 for x = 1, 2, 3, 4.

(a) Show that k = 2.5. [2]

(b) Find E(X). [2]

Markscheme

Part	Answer	Marks
(a)	(10 + 4k)/20 = 1	M1
	k = 2.5	AG A1
(b)	Σx(x + 2.5)/20 = (30 + 25)/20	M1
	2.75	A1
(c)	(6.5/20)2	M1
	0.106	A1

Examiner's comments. Part (c) uses independence: multiply.

Question 2 (Paper 2 style) (about 15 minutes)

On average 8% of eggs in a supermarket are cracked. Eggs are sold in boxes of 12. Assume the eggs are cracked independently.

(a) Find the probability that a box contains no cracked eggs. [2]

(b) Find the probability that a box contains more than 2 cracked eggs. [2]

(d) A shopper checks 10 boxes. Find the probability that exactly 4 of them contain no cracked eggs. [3]

Markscheme

Part	Answer	Marks
(a)	0.9212	M1
	0.368	A1
(b)	1 − P(X ≤ 2)	M1
	0.0652	A1
(c)	0.96	A1
(d)	Y ~ B(10, 0.3677)	M1
	P(Y = 4)	M1
	0.245	A1

Examiner's comments. Part (d) uses the answer to (a) as the new p.

22 Common misconceptions and links

Students often think…	Actually…
E(X) must be a possible value	it is a long-run average
P(X ≥ 5) = 1 − P(X ≤ 5)	1 − P(X ≤ 4)
any repeated experiment is binomial	check all four conditions
Var(X) = np	np(1 − p)

How 4.7 connects to the rest of the course

Topic	Connection
1.6 Binomial theorem	nCr coefficients
4.5 Probability	independence
4.9 Normal distribution	continuous models
4.14 Continuous RVs (HL)	variance and E(aX + b)

23 Check your understanding

Section A: multiple choice

1. ΣP(X = x) =

A. 1 B. 0 C. E(X) D. n

2. A fair game has expected gain

A. 1 B. positive C. 0 D. the stake

3. E(X) for B(20, 0.4) is

A. 4.8 B. 8 C. 20 D. 0.4

4. Var(X) for B(20, 0.4) is

A. 8 B. 2.19 C. 0.24 D. 4.8

5. P(X > 3) =

A. 1 − P(X ≤ 2) B. 1 − P(X ≤ 3) C. P(X ≤ 3) D. 1 − P(X = 3)

Section B: structured

6. State the conditions for X ~ B(n, p). [2]

7. X ~ B(15, 0.2). Find P(X = 3) and P(X ≤ 3). [2]

Answers

1. A 2. C 3. B 4. D 5. B

6. Fixed n; two outcomes; constant p; independent trials.

7. 0.250; 0.648.

24 Summary and topic map

Idea	Remember
Distribution	ΣP = 1, 0 ≤ P ≤ 1
Expected value	E(X) = ΣxP(X = x)
Fair game	expected gain 0
Binomial conditions	fixed n, two outcomes, constant p, independent
Binomial formula	nCr pr(1 − p)n−r
Mean and variance	np; np(1 − p)

Before the exam, make sure you…

•define X and state its distribution
•convert inequalities for the cdf
•check that probabilities sum to 1
•use expectation for fairness

Discrete Random Variables and the Binomial on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 2. Topic map for Discrete Random Variables and the Binomial.

25 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. k = 0.3; 0.55

2. 0.14

Practice exercise 2

1. 3.5

2. 0

Practice exercise 3

1. (300 + 250)/500 − 2 = −$0.90

2. w/6 + 1/3 = 3: w = $16

Practice exercise 4

1. 70/256 = 0.273

2. 0.391; 0.158

3. 0.206

Practice exercise 5

1. 1 − p = 0.7: p = 0.3, n = 20

2. 3

Practice exercise 6

1. n = 26

2. 6

Practice exercise 7

1. 1/8, 3/8, 3/8, 1/8

2. 70/36 = 1.94

Practice exercise 8

1. 3.8

2. 0.2 + 0.15 + 0.133 + 0.025 = 0.508

Practice exercise 9

1. 91/6 − 12.25 = 2.92

Practice exercise 10

1. k = 1/12; E(X) = (3 + 8 + 15)/12 = 13/6

Practice exercise 11

1. 40/52 + 60/52 = $1.92

Practice exercise 12

1. B(4, 0.5) assuming independent births with equal probability

Practice exercise 13

1. 0.167

2. 0.122; 0.323

Practice exercise 14

1. 4; 1.92

Practice exercise 15

1. 45

Practice exercise 16

1. 2

Practice exercise 17

1. 1 − (1 − 0.0302)5 = 0.142

Practice exercise 18

1. 0; 8

DP MATHEMATICS AA · TOPIC 4: STATISTICS AND PROBABILITY

4.9

The Normal Distribution

Full notes for SL · Syllabus SL 4.9, 4.12

Contents

1	Continuous data and the normal distribution
2	Properties of the normal curve
3	Calculating normal probabilities
4	Inverse normal calculations
5	Standardization and z-scores
6	Finding an unknown mean or standard deviation
7	Applications and combining with the binomial
8	When is a normal model appropriate?
9	Using symmetry
10	Probabilities between values
11	Inverse normal: one tail
12	Inverse normal: symmetric intervals
13	Standardizing with z-scores
14	Unknown mean or standard deviation
15	Unknown mean and standard deviation
16	Normal and binomial combined
17	Normal distribution in quality control
18	Mixed review: worked problems
19	Additional exam-style practice
20	The 68–95–99.7 rule in estimation
21	Technology checklist for the normal distribution
22	Exam-style questions with markscheme
23	Common misconceptions and links
24	Check your understanding
25	Summary and topic map
26	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
recognize situations modelled by a normal distribution
describe the properties of the normal curve, including the 68–95–99.7 rule
calculate normal probabilities with technology
find values from given probabilities (inverse normal)
standardize values using z = (x − μ)/σ and interpret z-scores
find an unknown mean or standard deviation
combine normal and binomial distributions in context

This booklet covers SL 4.9, 4.12, which is common to SL and HL.

Guiding questions

Why do so many natural measurements follow a bell-shaped curve? How can probabilities for continuous quantities be calculated?

How this topic is examined

The normal distribution appears in almost every Paper 2, often combined with another topic: find a probability, find a value from a percentage (inverse normal), find an unknown mean or standard deviation using z-scores, and then use a binomial distribution for repeated selections.

Sketch a normal curve and shade the required region for every question; it prevents errors with 'less than' and 'greater than'.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Continuous data and the normal distribution

The bell curve.

Think about it

If you measured the heights of 10 000 adults and drew a histogram with very narrow classes, what shape would you expect?

Many continuous measurements (heights, masses, measurement errors, test scores) cluster around a central value, with values further away becoming steadily less common. They are modelled by the normal distribution, X ~ N(μ, σ2), with mean μ and standard deviation σ (variance σ2).

For a continuous random variable, probabilities are areas under the curve. The probability of any single exact value is 0, so P(X < a) = P(X ≤ a).

2 Properties of the normal curve

Symmetric about the mean.

The normal curve is bell-shaped and symmetric about μ; the mean, median and mode are equal. The total area under the curve is 1. The curve approaches but never reaches the x-axis. Increasing σ makes the curve wider and flatter.

The 68–95–99.7 rule: approximately 68% of values lie within 1σ of the mean, 95% within 2σ, and 99.7% within 3σ.

Figure 1. About 68%, 95% and 99.7% of values lie within 1, 2 and 3 standard deviations of the mean.

WORKED EXAMPLE 1 Using the rule (basic)

Heights are normally distributed with mean 170 cm and SD 8 cm. Estimate the percentage of heights between 154 and 186 cm, and above 178 cm.

Solution

154 to 186 is μ ± 2σ: about 95%A1

Above μ + σ: (100 − 68)/2 = 16%A1

3 Calculating normal probabilities

Use the GDC normal cdf.

For X ~ N(μ, σ2), the GDC normal cdf gives P(a < X < b) from the lower bound, upper bound, μ and σ. For an open interval use a very large or very small bound (such as ±1099). Note that the GDC needs σ, not σ2.

WORKED EXAMPLE 2 Probabilities (standard)

The mass of apples is X ~ N(150, 122) grams. Find (a) P(X < 140); (b) P(X > 175); (c) P(130 < X < 160).

Solution

(a) 0.202A1

(b) 0.0186A1

Figure 2. P(130 < X < 160) for X ~ N(150, 122).

Common mistake

Entering the variance instead of the standard deviation: N(150, 144) has σ = 12, not 144.

Practice exercise 1

1. X ~ N(50, 42). Find P(X > 57) and P(45 < X < 52).

2. Times to run 1 km are N(5.2, 0.42) minutes. What percentage take longer than 6 minutes?

3. Using symmetry, show that P(X < μ − a) = P(X > μ + a).

4 Inverse normal calculations

From a probability to a value.

Given a probability (area), the inverse normal function finds the value x such that P(X < x) equals that area. For an upper tail, P(X > x) = p, use the area 1 − p to the left (or set the tail option to 'right' on calculators that allow it).

WORKED EXAMPLE 3 Inverse normal (standard)

For the apples, find (a) the mass exceeded by 10% of apples; (b) the interquartile range.

Solution

(a) P(X < m) = 0.9: m = 165 g (165.4)A1

(b) Q1 = 141.9, Q3 = 158.1M1

IQR = 16.2 gA1

Practice exercise 2

1. X ~ N(80, 152). Find the value of k for which P(X < k) = 0.3.

2. Exam scores are N(62, 112). The top 5% get an A. Find the minimum score for an A.

5 Standardization and z-scores

Measure in standard deviations.

The z-score of a value tells you how many standard deviations it lies from the mean: positive above the mean, negative below. Every normal distribution transforms into the standard normal Z ~ N(0, 1). z-scores allow comparisons between different distributions.

WORKED EXAMPLE 4 Comparing performance (standard)

Ana scored 72 in a test with mean 60 and SD 8; Ben scored 81 in a test with mean 70 and SD 10. Who performed better relative to their group?

Solution

Ana: z = (72 − 60)/8 = 1.5A1

Ben: z = (81 − 70)/10 = 1.1A1

Ana performed better relative to her groupR1

Practice exercise 3

1. X ~ N(40, 52). Find the z-score of 32.

2. A value has z-score 2.4 in N(100, 152). Find the value.

6 Finding an unknown mean or standard deviation

Use z-scores and inverse normal.

If μ or σ is unknown, use the inverse normal on Z ~ N(0, 1) to find the z-value for the given probability, then solve (x − μ)/σ = z. With both unknown, two conditions give two simultaneous equations.

WORKED EXAMPLE 5 Unknown standard deviation (exam level)

X ~ N(80, σ2) and P(X > 92) = 0.15. Find σ.

Solution

P(Z < z) = 0.85 gives z = 1.0364M1

(92 − 80)/σ = 1.0364M1

σ = 11.6A1

WORKED EXAMPLE 6 Both unknown (exam level)

X ~ N(μ, σ2) with P(X < 30) = 0.2 and P(X > 50) = 0.1. Find μ and σ.

Solution

z-values: −0.8416 and 1.2816A1

μ − 0.8416σ = 30 and μ + 1.2816σ = 50M1

2.1232σ = 20: σ = 9.42, μ = 37.9A1 A1

Practice exercise 4

1. X ~ N(μ, 62) and P(X < 40) = 0.25. Find μ.

2. Bags of sugar have mass N(μ, 82) g. The machine is set so that only 2% of bags weigh less than 500 g. Find μ.

7 Applications and combining with the binomial

Repeated selections.

A common exam structure: use the normal distribution to find a probability p for one item, then use X ~ B(n, p) for the number of items (out of n chosen independently) that satisfy the condition.

WORKED EXAMPLE 7 Normal then binomial (exam level)

Apple masses are N(150, 122) g. An apple is 'large' if it weighs over 165 g. In a bag of 10 apples, find the probability that at least 2 are large.

Solution

p = P(X > 165) = 0.1056A1

Y ~ B(10, 0.1056)M1

P(Y ≥ 2) = 1 − P(Y ≤ 1) = 0.286A1

Practice exercise 5

1. Battery lifetimes are N(400, 302) hours. Find the probability that exactly 3 of 5 batteries last more than 420 hours.

8 When is a normal model appropriate?

Checking the shape.

A normal model suits continuous data that are roughly symmetric and bell-shaped, with most values near the mean and fewer further away. Check with a histogram. Data that are skewed, bounded close to the mean (such as times that cannot be negative with a large SD), or bimodal are poorly modelled.

WORKED EXAMPLE 8 Checking (standard)

Waiting times have mean 4 min and SD 3 min. Comment on a normal model.

Solution

P(T < 0) = P(Z < −1.33) = 0.0912 is not negligibleM1 A1

Times cannot be negative: the model is not appropriateR1

Practice exercise 6

1. Give two examples of variables that are approximately normal.

2. Why might exam scores out of 20 not be normal?

9 Using symmetry

Areas without a calculator.

The normal curve is symmetric about μ, so P(X < μ − a) = P(X > μ + a) and P(X < μ) = 0.5. These facts let you find probabilities from given information without a GDC.

WORKED EXAMPLE 9 Symmetry (standard)

X ~ N(50, σ2) and P(X > 58) = 0.15. Find P(42 < X < 58).

Solution

P(X < 42) = 0.15M1

P(42 < X < 58) = 1 − 0.3 = 0.7A1

Practice exercise 7

1. X ~ N(20, σ2) and P(X < 17) = 0.2. Find P(X > 23) and P(20 < X < 23).

10 Probabilities between values

Normal cdf with two bounds.

P(a < X < b) is the area between a and b. With a GDC, enter lower bound a, upper bound b, μ and σ. Sketch and shade the region to check that the answer is sensible.

Figure 3. P(160 < X < 185) for X ~ N(170, 92).

WORKED EXAMPLE 10 Between values (standard)

Heights are N(170, 92) cm. Find P(160 < X < 185).

Solution

normalcdf(160, 185, 170, 9)M1

= 0.819A1

Practice exercise 8

1. Find P(|X − 170| > 15).

2. Find the expected number of 500 people with heights between 160 and 185 cm.

11 Inverse normal: one tail

Finding a value.

The inverse normal function returns the value x with P(X < x) = p. For a right tail, P(X > x) = p, use 1 − p as the left area (or the 'right' tail option).

WORKED EXAMPLE 11 Top 10% (standard)

Find the height exceeded by 10% of people (N(170, 92)).

Solution

P(X < x) = 0.9M1

x = 181.5 cmA1

Practice exercise 9

1. Find the 25th percentile of N(170, 92).

2. Find the height exceeded by 99% of people.

12 Inverse normal: symmetric intervals

Middle percentages.

For an interval containing the middle p% of values, each tail has (100 − p)/2 %. Use the inverse normal for the lower tail and symmetry for the upper bound.

WORKED EXAMPLE 12 Middle 90% (standard)

Find the interval containing the middle 90% of N(170, 92).

Solution

Lower 5%: 155.2; upper 95%: 184.8M1 A1

155.2 < X < 184.8A1

Practice exercise 10

1. Find k such that P(170 − k < X < 170 + k) = 0.95.

13 Standardizing with z-scores

Comparing different distributions.

z = (x − μ)/σ measures how many standard deviations x is from the mean. Z ~ N(0, 1). z-scores allow comparisons between different scales and are needed to find unknown parameters.

WORKED EXAMPLE 13 Comparison (standard)

A student scores 68 in Physics (μ = 60, σ = 5) and 75 in Chemistry (μ = 70, σ = 4). Which result is better relative to the class?

Solution

z = 1.6 and 1.25A1 A1

PhysicsR1

Practice exercise 11

1. Find the value with z-score −0.8 in N(90, 122).

14 Unknown mean or standard deviation

From one condition.

If one parameter is unknown, find the z-value for the given probability with inverse normal on N(0, 1), then solve (x − μ)/σ = z.

WORKED EXAMPLE 14 Unknown mean (exam level)

X ~ N(μ, 62) and P(X < 40) = 0.1. Find μ.

Solution

z = −1.2816A1

μ = 40 + 1.2816 × 6M1

μ = 47.7A1

Practice exercise 12

1. X ~ N(100, σ2) and P(X > 120) = 0.05. Find σ.

15 Unknown mean and standard deviation

Two conditions.

Two probability conditions give two equations μ + z1σ = x1 and μ + z2σ = x2. Solve simultaneously.

WORKED EXAMPLE 15 Both unknown (exam level)

P(X < 30) = 0.15 and P(X > 60) = 0.05. Find μ and σ.

Solution

z-values −1.0364 and 1.6449A1

μ − 1.0364σ = 30, μ + 1.6449σ = 60M1

σ = 11.19, μ = 41.6A1 A1

Practice exercise 13

1. 10% of values are below 20 and 20% above 50. Find μ and σ.

16 Normal and binomial combined

One item, then many.

First use the normal distribution to find p, the probability that one item satisfies a condition. Then the number of items (out of n independent ones) satisfying it is B(n, p).

WORKED EXAMPLE 16 Combined (exam level)

Eggs have mass N(62, 42) g; 'large' means over 66 g. Find P(at least 3 large eggs in a box of 12).

Solution

p = P(X > 66) = 0.1587A1

Y ~ B(12, 0.1587): P(Y ≥ 3) = 0.294M1 A1

Practice exercise 14

1. Find the expected number of large eggs in a box of 12.

17 Normal distribution in quality control

Specifications and tolerances.

Manufacturers set a target value and tolerance. The proportion out of specification is the area outside the tolerance limits. Adjusting the mean or reducing the SD reduces this proportion.

WORKED EXAMPLE 17 Out of specification (exam level)

Rods have length N(50.0, 0.22) mm and must be between 49.6 and 50.4 mm. Find the proportion rejected, and the SD needed to reject only 1%.

Solution

P(outside) = 1 − 0.9545 = 0.0455M1 A1

Need 0.4/σ = 2.5758: σ = 0.155 mmM1 A1

Practice exercise 15

1. If the mean drifts to 50.1 mm (SD 0.2), find the proportion rejected.

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 18 Mixed problem (exam level)

Journey times are N(42, 52) min. (a) Find P(time > 50). (b) Find the time t such that 80% of journeys take less than t. (c) Over 10 journeys, find P(exactly 2 take more than 50 min).

Solution

(a) 0.0548A1

(b) 46.2 minA1

Practice exercise 16

1. Find the IQR of N(42, 52).

19 Additional exam-style practice

Paper 2.

WORKED EXAMPLE 19 Paper 2 question (exam level)

The lifetime of a battery is N(μ, 402) hours, and 20% last less than 450 hours. (a) Find μ. (b) Find P(a battery lasts more than 520 hours). (c) A pack has 4 batteries. Find P(all last more than 450 hours).

Solution

(a) z = −0.8416: μ = 483.7M1 A1

(b) 0.182A1

Practice exercise 17

1. Find the median lifetime.

20 The 68–95–99.7 rule in estimation

Quick mental checks.

About 68% of values lie within 1σ, 95% within 2σ and 99.7% within 3σ of the mean. Use these to estimate probabilities quickly and to check GDC answers.

WORKED EXAMPLE 20 Estimate (standard)

X ~ N(100, 152). Estimate P(X > 130) and P(85 < X < 115).

Solution

130 is 2σ above: about 2.5%A1

Within 1σ: about 68%A1

Practice exercise 18

1. Estimate P(X < 55) for N(100, 152).

21 Technology checklist for the normal distribution

Avoiding common errors.

Use σ, not σ2. Use a large number such as 1099 for an infinite bound. For inverse normal, enter the area to the left (unless using the tail option). Write down the distribution and the values entered.

WORKED EXAMPLE 21 Check (standard)

X ~ N(25, 16). A student enters σ = 16 to find P(X < 30). Explain the error and correct it.

Solution

σ = 4, not 16R1

P(X < 30) = 0.894A1

GDC tip

normalcdf(lower, upper, μ, σ) for probabilities; invNorm(area, μ, σ) for values; invNorm(area, 0, 1) for z-values when μ or σ is unknown.

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 2 style) (about 15 minutes)

The time T for a delivery is normally distributed with mean 42 minutes and standard deviation 6 minutes.

(a) Find the probability that a delivery takes more than 50 minutes. [2]

(b) Find the time exceeded by 20% of deliveries. [2]

(c) A delivery is 'late' if it takes more than 50 minutes. Find the probability that, of 8 deliveries, at most 1 is late. [3]

(d) The company changes its process so that the mean remains 42 minutes but only 5% of deliveries take longer than 50 minutes. Find the new standard deviation. [3]

Markscheme

Part	Answer	Marks
(a)	normal cdf	M1
	0.0912	A1
(b)	inverse normal with 0.8	M1
	47.0 min	A1
(c)	B(8, 0.0912)	M1
	P(Y ≤ 1)	M1
	0.839	A1
(d)	z = 1.6449	A1
	8/σ = 1.6449	M1
	σ = 4.86	A1

Examiner's comments. Use unrounded probabilities in part (c).

23 Common misconceptions and links

Students often think…	Actually…
enter σ2 into the GDC	enter σ
P(X = a) > 0 for continuous X	it is 0
a negative z-score is an error	it means below the mean
all data are normal	check the shape is symmetric and unimodal

How 4.9 connects to the rest of the course

Topic	Connection
4.1 Statistics	mean and standard deviation
4.7 Binomial	combining distributions
4.14 Continuous RVs (HL)	probability density functions
5.5 Integration	areas under curves

24 Check your understanding

Section A: multiple choice

1. About what percentage lies within 2σ of μ?

A. 95% B. 68% C. 99.7% D. 50%

2. For N(μ, σ2), the median is

A. σ B. 0 C. μ D. μ + σ

3. z = (x − μ)/σ measures

A. percentiles B. standard deviations from the mean C. variance D. frequency

4. P(X > μ) =

A. 1 B. 0 C. 0.68 D. 0.5

5. Inverse normal finds

A. a probability from a value B. a value from a probability C. the mean D. σ2

Section B: structured

6. X ~ N(20, 32). Find P(|X − 20| < 4). [2]

7. Find k such that P(μ − kσ < X < μ + kσ) = 0.9. [2]

Answers

1. A 2. C 3. B 4. D 5. B

6. 0.818

7. k = 1.645

25 Summary and topic map

Idea	Remember
Model	X ~ N(μ, σ2); symmetric bell curve
68–95–99.7	within 1, 2, 3 standard deviations
Probabilities	normal cdf (areas)
Inverse normal	value from a probability
z-scores	z = (x − μ)/σ
Unknown μ or σ	inverse normal on Z, then solve
Combined	normal p, then binomial

Before the exam, make sure you…

•sketch and shade
•enter σ not σ2
•use the correct tail for inverse normal
•keep probabilities unrounded

The Normal Distribution on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 4. Topic map for The Normal Distribution.

26 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. 0.0401; 0.586

2. 2.28%

3. the curve is symmetric about μ

Practice exercise 2

1. 72.1

2. 80.1

Practice exercise 3

1. −1.6

2. 136

Practice exercise 4

1. 44.0

2. 516.4 g

Practice exercise 5

1. p = 0.2525; B(5, 0.2525): P(Y = 3) = 0.0899

Practice exercise 6

1. heights, measurement errors

2. bounded between 0 and 20, possibly skewed

Practice exercise 7

1. 0.2; 0.3

Practice exercise 8

1. 0.0956

2. 409

Practice exercise 9

1. 163.9 cm

2. 149.1 cm

Practice exercise 10

1. k = 17.6

Practice exercise 11

1. 80.4

Practice exercise 12

1. 12.2

Practice exercise 13

1. σ = 14.1, μ = 38.1

Practice exercise 14

1. 1.90

Practice exercise 15

1. P(X < 49.6) + P(X > 50.4) = 0.0062 + 0.0668 = 0.0730

Practice exercise 16

1. 6.74

Practice exercise 17

1. 483.7 hours

Practice exercise 18

1. 3σ below: about 0.15%

revisionprep.comIB-style notes · question bank · mock papers